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Suppose we have the points $a_1,a_2,\dots, a_{42}$ equally spaced around a circle. In how many ways can we choose exactly three points so that they form the vertices of an isosceles (or equilateral) triangle? (Of course, this implies permutations of triples are counted as the same)

I have tried counting the number of possibilities by fixing one vertex, then finding the other two based on the fixed point. In this manner, we may find the probability, then multiply by the number of events. I found the probability, by fixing one vertex, to be 3/166. Then, multiplying by 42*41*40, I have received an answer of $2520$ in this manner which I am sure is an over counting. what is a way to solve this problem then?

Thank you!

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Putting vertices of triangle aside, let's call $a$ the number of points on the circle near by each of the equal sides and $b$ the number of points beside the other side. We have $2a+b=42-3=39$. This equation has 20 non-negative integer answers, and one of them $a=13,b=13$ is an equilateral triangle. (all sides equal). We can rotate the non-equilateral ones around the circle each 42 times, and the equilateral one $42/3$ times so at last we get $42*19+14=812$

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  • $\begingroup$ I edited my answer to correct my mistake in calculation, I mistakenly wrote $42/3=21$ while it is $14$! $\endgroup$ – hhsaffar Nov 28 '13 at 17:16
  • $\begingroup$ I don't get what you mean by : "the number of points on the circle near by each of the equal sides and b the number of points beside the other side". Could you explain a bit more? Also, how did you get that the equation has 20 integer solutions? $\endgroup$ – 1110101001 Dec 4 '13 at 0:01
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Count the equilateral and the isosceles non-equilateral separately.

For the equilateral, one of the vertices is $a_1$ to $a_{14}$, and then the rest are determined.

For the non-equilateral isosceles triangles $ABC$, the vertex $A$ such that $AB=AC$ is special. It can be picked in $42$ ways. For each of these ways, count the number of ways to choose the other two vertices. You can assume for that part of the counting that your special vertex is $a_1$, and that the second one is less than halfway along. That is the only one you need to pick, since then the third vertex is determined. Don't forget to not count the equilateral case.

After you do the arithmetic, you should end up with $812$.

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    $\begingroup$ Oh, I devided $43$ by $3$ and wrote $21$! You helped me find my mistake! $\endgroup$ – hhsaffar Nov 28 '13 at 17:25

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